Skip to Content
OpenStax Logo
University Physics Volume 1

9.5 Collisions in Multiple Dimensions

University Physics Volume 19.5 Collisions in Multiple Dimensions
Buy book
  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

Learning Objectives

By the end of this section, you will be able to:
  • Express momentum as a two-dimensional vector
  • Write equations for momentum conservation in component form
  • Calculate momentum in two dimensions, as a vector quantity

It is far more common for collisions to occur in two dimensions; that is, the angle between the initial velocity vectors is neither zero nor 180°180°. Let’s see what complications arise from this.

The first idea we need is that momentum is a vector; like all vectors, it can be expressed as a sum of perpendicular components (usually, though not always, an x-component and a y-component, and a z-component if necessary). Thus, when we write down the statement of conservation of momentum for a problem, our momentum vectors can be, and usually will be, expressed in component form.

The second idea we need comes from the fact that momentum is related to force:

F=dpdt.F=dpdt.

Expressing both the force and the momentum in component form,

Fx=dpxdt,Fy=dpydt,Fz=dpzdt.Fx=dpxdt,Fy=dpydt,Fz=dpzdt.

Remember, these equations are simply Newton’s second law, in vector form and in component form. We know that Newton’s second law is true in each direction, independently of the others. It follows therefore (via Newton’s third law) that conservation of momentum is also true in each direction independently.

These two ideas motivate the solution to two-dimensional problems: We write down the expression for conservation of momentum twice: once in the x-direction and once in the y-direction.

pf,x=p1,i,x+p2,i,xpf,y=p1,i,y+p2,i,ypf,x=p1,i,x+p2,i,xpf,y=p1,i,y+p2,i,y
(9.18)

This procedure is shown graphically in Figure 9.22.

Figure a, titled break initial momentum into x and y components shows vector p 1 i as a solid arrow pointing to the right and down. Its components are shown as dashed arrows: p 1 i y points down from the tail of p 1 i and p 1 i x points to the right from the head of p 1 i y to the head of p 1 i. Vector p 2 i is shown as a solid arrow with its tail at the head of vector p 1 i, and is shorter than p 1 i. Vector p 2 i points to the right and up. Its components are shown as dashed arrows: p 2 i x points to the right from the tail of p 2 i and p 2 i y points up from the head of p 2 i x to the head of p 2 i. Vector p f points from the tail of p 1 i to the head of p 2 i, pointing to the right and slightly down. Figure b titled add x and y components to obtain x and y components of final momentum shows the vector sums of the components. P 1 i y is a downward arrow. P 2 i y is a shorter upward arrow, aligned with its tail at the head of P 1 i y. P f y is a short downward arrow that starts at the tail of P 1 i y and ends at the head of P 2 i y. P 1 i x is a rightward arrow. P 2 i x is a shorter rightward arrow, aligned with its tail at the head of P 1 i x. P f x is a long rightward arrow that starts at the tail of P 1 i x and ends at the head of P 2 i x. Figure c, titled add x and y components of final momentum shows the right triangle formed by sides p f x and p f y and hypotenuse p f. Arrows from figure b indicate that p f x and p f y are the same in figures b and c.
Figure 9.22 (a) For two-dimensional momentum problems, break the initial momentum vectors into their x- and y-components. (b) Add the x- and y-components together separately. This gives you the x- and y-components of the final momentum, which are shown as red dashed vectors. (c) Adding these components together gives the final momentum.

We solve each of these two component equations independently to obtain the x- and y-components of the desired velocity vector:

vf,x=m1v1,i,x+m2v2,i,xmvf,y=m1v1,i,y+m2v2,i,ym.vf,x=m1v1,i,x+m2v2,i,xmvf,y=m1v1,i,y+m2v2,i,ym.

(Here, m represents the total mass of the system.) Finally, combine these components using the Pythagorean theorem,

vf=|vf|=vf,x2+vf,y2.vf=|vf|=vf,x2+vf,y2.

Problem-Solving Strategy: Conservation of Momentum in Two Dimensions

The method for solving a two-dimensional (or even three-dimensional) conservation of momentum problem is generally the same as the method for solving a one-dimensional problem, except that you have to conserve momentum in both (or all three) dimensions simultaneously:

  1. Identify a closed system.
  2. Write down the equation that represents conservation of momentum in the x-direction, and solve it for the desired quantity. If you are calculating a vector quantity (velocity, usually), this will give you the x-component of the vector.
  3. Write down the equation that represents conservation of momentum in the y-direction, and solve. This will give you the y-component of your vector quantity.
  4. Assuming you are calculating a vector quantity, use the Pythagorean theorem to calculate its magnitude, using the results of steps 3 and 4.

Example 9.14

Traffic Collision A small car of mass 1200 kg traveling east at 60 km/hr collides at an intersection with a truck of mass 3000 kg that is traveling due north at 40 km/hr (Figure 9.23). The two vehicles are locked together. What is the velocity of the combined wreckage?

An x y coordinate system is shown. A large truck mass m T = 3000 kilograms is moving north toward the origin with velocity v T. A small car mass m c = 1200 kilograms is moving east toward the origin with velocity v c, which is less than v T.
Figure 9.23 A large truck moving north is about to collide with a small car moving east. The final momentum vector has both x- and y-components.

Strategy First off, we need a closed system. The natural system to choose is the (car + truck), but this system is not closed; friction from the road acts on both vehicles. We avoid this problem by restricting the question to finding the velocity at the instant just after the collision, so that friction has not yet had any effect on the system. With that restriction, momentum is conserved for this system.

Since there are two directions involved, we do conservation of momentum twice: once in the x-direction and once in the y-direction.

Solution Before the collision the total momentum is

p=mcvc+mTvT.p=mcvc+mTvT.

After the collision, the wreckage has momentum

p=(mc+mT)vw.p=(mc+mT)vw.

Since the system is closed, momentum must be conserved, so we have

mcvc+mTvT=(mc+mT)vw.mcvc+mTvT=(mc+mT)vw.

We have to be careful; the two initial momenta are not parallel. We must add vectorially (Figure 9.24).

Arrow p c points horizontally to the right. Arrow p t points vertically upward. The head of p t meets the tail of p c. P t is longer than p t. A dashed line is shown from the tail of p t to the head of p c. The angle between the dashed line and p t, at the tail of p t, is labeled as theta.
Figure 9.24 Graphical addition of momentum vectors. Notice that, although the car’s velocity is larger than the truck’s, its momentum is smaller.

If we define the +x-direction to point east and the +y-direction to point north, as in the figure, then (conveniently),

pc=pci^=mcvci^pT=pTj^=mTvTj^.pc=pci^=mcvci^pT=pTj^=mTvTj^.

Therefore, in the x-direction:

mcvc=(mc+mT)vw,xvw,x=(mcmc+mT)vcmcvc=(mc+mT)vw,xvw,x=(mcmc+mT)vc

and in the y-direction:

mTvT=(mc+mT)vw,yvw,y=(mTmc+mT)vT.mTvT=(mc+mT)vw,yvw,y=(mTmc+mT)vT.

Applying the Pythagorean theorem gives

|vw|=[(mcmc+mt)vc]2+[(mtmc+mt)vt]2=[(1200kg4200kg)(16.67ms)]2+[(3000kg4200kg)(11.1ms)]2=(4.76ms)2+(7.93ms)2=9.25ms33.3kmhr.|vw|=[(mcmc+mt)vc]2+[(mtmc+mt)vt]2=[(1200kg4200kg)(16.67ms)]2+[(3000kg4200kg)(11.1ms)]2=(4.76ms)2+(7.93ms)2=9.25ms33.3kmhr.

As for its direction, using the angle shown in the figure,

θ=tan−1(vw,xvw,y)=tan−1(7.93m/s4.76m/s)=59°.θ=tan−1(vw,xvw,y)=tan−1(7.93m/s4.76m/s)=59°.

This angle is east of north, or 31°31° counterclockwise from the +x-direction.

Significance As a practical matter, accident investigators usually work in the “opposite direction”; they measure the distance of skid marks on the road (which gives the stopping distance) and use the work-energy theorem along with conservation of momentum to determine the speeds and directions of the cars prior to the collision. We saw that analysis in an earlier section.

Check Your Understanding 9.9

Suppose the initial velocities were not at right angles to each other. How would this change both the physical result and the mathematical analysis of the collision?

Example 9.15

Exploding Scuba Tank A common scuba tank is an aluminum cylinder that weighs 31.7 pounds empty (Figure 9.25). When full of compressed air, the internal pressure is between 2500 and 3000 psi (pounds per square inch). Suppose such a tank, which had been sitting motionless, suddenly explodes into three pieces. The first piece, weighing 10 pounds, shoots off horizontally at 235 miles per hour; the second piece (7 pounds) shoots off at 172 miles per hour, also in the horizontal plane, but at a 19°19° angle to the first piece. What is the mass and initial velocity of the third piece? (Do all work, and express your final answer, in SI units.)

A drawing of a scuba tank exploding, and the resulting three pieces of different sizes.
Figure 9.25 A scuba tank explodes into three pieces.

StrategyTo use conservation of momentum, we need a closed system. If we define the system to be the scuba tank, this is not a closed system, since gravity is an external force. However, the problem asks for just the initial velocity of the third piece, so we can neglect the effect of gravity and consider the tank by itself as a closed system. Notice that, for this system, the initial momentum vector is zero.

We choose a coordinate system where all the motion happens in the xy-plane. We then write down the equations for conservation of momentum in each direction, thus obtaining the x- and y-components of the momentum of the third piece, from which we obtain its magnitude (via the Pythagorean theorem) and its direction. Finally, dividing this momentum by the mass of the third piece gives us the velocity.

Solution First, let’s get all the conversions to SI units out of the way:

31.7lb×1kg2.2lb14.4kg 10lb4.5kg 235mileshour×1hour3600s×1609mmile=105ms 7lb3.2kg 172milehour=77ms m3=14.4kg(4.5kg+3.2kg)=6.7kg.31.7lb×1kg2.2lb14.4kg 10lb4.5kg 235mileshour×1hour3600s×1609mmile=105ms 7lb3.2kg 172milehour=77ms m3=14.4kg(4.5kg+3.2kg)=6.7kg.

Now apply conservation of momentum in each direction.

The three pieces of the scuba tank are shown on an x y coordinate system. The medium size piece is on the positive x axis and has momentum p 1 in the plus x direction. The smallest piece is at an angle theta above the positive x axis and has momentum p 2. The largest piece is at an angle phi below the negative x axis and has momentum p 3.

x-direction:

pf,x=p0,xp1,x+p2,x+p3,x=0m1v1,x+m2v2,x+p3,x=0p3,x=m1v1,xm2v2,xpf,x=p0,xp1,x+p2,x+p3,x=0m1v1,x+m2v2,x+p3,x=0p3,x=m1v1,xm2v2,x

y-direction:

pf,y=p0,yp1,y+p2,y+p3,y=0m1v1,y+m2v2,y+p3,y=0p3,y=m1v1,ym2v2,ypf,y=p0,yp1,y+p2,y+p3,y=0m1v1,y+m2v2,y+p3,y=0p3,y=m1v1,ym2v2,y

From our chosen coordinate system, we write the x-components as

p3,x=m1v1m2v2cosθ=(14.5kg)(105ms)(4.5kg)(77ms)cos(19°)=−1850kg·ms.p3,x=m1v1m2v2cosθ=(14.5kg)(105ms)(4.5kg)(77ms)cos(19°)=−1850kg·ms.

For the y-direction, we have

p3y=0m2v2sinθ=(4.5kg)(77ms)sin(19°)=−113kg·ms.p3y=0m2v2sinθ=(4.5kg)(77ms)sin(19°)=−113kg·ms.

This gives the magnitude of p3p3:

p3=p3,x2+p3,y2=(−1850kg·ms)2+(−113kg·ms)=1854kg·ms.p3=p3,x2+p3,y2=(−1850kg·ms)2+(−113kg·ms)=1854kg·ms.

The velocity of the third piece is therefore

v3=p3m3=1854kg·ms6.7kg=277ms.v3=p3m3=1854kg·ms6.7kg=277ms.

The direction of its velocity vector is the same as the direction of its momentum vector:

ϕ=tan−1(p3,yp3,x)=tan−1(113kg·ms1850kg·ms)=3.5°.ϕ=tan−1(p3,yp3,x)=tan−1(113kg·ms1850kg·ms)=3.5°.

Because ϕϕ is below the xx-axis, the actual angle is 183.5°183.5° from the +x-direction.

Significance The enormous velocities here are typical; an exploding tank of any compressed gas can easily punch through the wall of a house and cause significant injury, or death. Fortunately, such explosions are extremely rare, on a percentage basis.

Check Your Understanding 9.10

Notice that the mass of the air in the tank was neglected in the analysis and solution. How would the solution method changed if the air was included? How large a difference do you think it would make in the final answer?

Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction
Citation information

© Sep 19, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.